Thesaurarium mathematicae, or, The treasury of mathematicks containing variety of usefull practices in arithmetick, geometry, trigonometry, astronomy, geography, navigation and surveying ... to which is annexed a table of 10000 logarithms, log-sines, and log-tangents / by John Taylor.

CHAP. IV. Of GEOMETRY.

* 1.1THE End and Scope of Geometry is to measure well: for as Number or Multitude, is the Subject of Arithmetick: so is Magnitude that of Geometry: to measure well is therefore to consider the Nature of e∣very thing that is to be measured; to compare such like things one with another: and to un∣derstand their Reason and proportion, and also their similitude: And this is the End and Scope of Geometry* 1.2.

I shall not trouble you with the Definitions of Geometry, they being largely handled by many, and herein every one meanly conver∣sant in the study of the Mathematicks is ac∣quainted,

but shall immediately fall in hand with the principal Propositions, which chiefly concern the other following parts of this treatise.

SECT. I. The Explication of some Geometrical Pro∣positions.

PROP. I. To erect a perpendicular on any part of a line assigned.

LET the Line be A, B, and on the point D,* 1.3 'tis required to raise a perpendicular to A, B, To operate which first open your Compasses to any convenient distance, and placing one foot thereof in D, with the other make the two marks C, and E, equidistant from D; then open the Compasses to some other convenient distance, and set one foot in E, and describe the Arch FF; then likewise in C, describe the Arch GG, then through the Intersections of these two Arches, and to the point D, draw H D, per∣pendicular to A B; as was required.

PROP. II. To Erect a Perpendicular, on the End of a Line.

Let the given line be A B, and on the End* 1.4 thereof at B, 'tis required to raise a Perpen∣dicular

line: To perform which open your Compasses to the distance B D, then on B as a Center, describe the Arch D, E, F, then from D, to E, place BD; then placing one foot in E, describe the Arch CF, then remove your* 1.5 Compasses to F, and draw the Arch CE; Lastly through their Intersection draw C B, which is a Perpendicular to AB, on the end B; as required.

PROP. III. From a Point above to let fall a Perpendicular on a Line.

Let the line given be B A, and 'tis required* 1.6 from the point above at C; to let fall a Per∣pendicular to the said Line: To perform which place one foot of your Compasses in C, and o∣pen them beyond the given line A B, and de∣scribe the Arch EF; divide EF, into two parts in D; Lastly draw CD, which shall be per∣pendicular unto AB, falling from the point above at C, as was so required.

PROP. IV. To draw a right line Parallel to a right line, at a∣ny distance assigned.

Let the distance assigned be O E, and the* 1.7 Lime given be A B, and 'tis required to draw C D, Parallel to A B; at the distance O E: To perform which, take in your Com∣passes the distance O E, and on A, describe

the Arch H, and on B, the Arch K; then draw C D, so as it may justly touch the two Arches, but cut them not, so shall C D; be pa∣rallel to A B, at the assigned distance O E, as was required.

PROP. V. To Protract an Angle of any Quantity of Degrees propounded.

Let it be required to Protract, or lay down an Angle, of 40 degrees: To perform which first draw a right line as A B, then open y••••r Compasses to 60 degrees, in your line of Chords: and with that Distance on A, de∣scribe* 1.8 the Arch E F, then take 40 degrees in your Compasses out of your line of Chords, and place it on the Arch, from F, to H; Lastly through the point H, and from A draw A C; so shall the Angle CAB contain 40 degrees as required.

PROP. VI. To measure an Angle already protracted.

Let the Angle given be C A B, and 'tis ••equired to find the Quantity thereof: To ••erform which take in your Compasses 60 de••rees from your line of Chords; and on A,* 1.9 ••escribe the Arch EF; then take in your Com∣••asses the Distance FH, and apply it to your 〈…〉〈…〉ne of Chords; and you will find the Angle, 〈…〉〈…〉 AB to contain 40 degrees.

PROP. VII. To divide an Angle into two Equal parts.

Let the Angle given be BAC, and 'tis required to divide it into two equal parts: To perform which do thus: first take in your Com∣passes any convenient distance, and placing one foot in A, describe the Arch FKHE,* 1.10 then on H, describe the Arch KK, and on K, the Arch HH; lastly through the Intersections of these two Arches, draw the line AD, to the Angular point A; so shall the Angle BAC, be divided into two equal parts, viz. BA••, and DAC; as required.

PROP. VIII. To divide a right line into any Number of Equal or Unequal parts; or like to any divided line propounded.

Let the line A B, be given to be divided into 5 equal parts; as the line CD. To per∣form which do thus: first on the point C, draw out a line making an Angle with CD at plea∣sure: then make CF, equal to AB; and joyn their Extremities FD, then draw Parallel lines* 1.11 to FD, through all the 5 points of CD, (by the 4 prop. aforegoing) which shall divide AB, into 5 equal parts; as required: This way is to be observed, when the line given to be divided, is greater than the divided line propounded.

CASE II.

But if AB, be shorter than the given di∣vided line CD; take the line AB, in your Compasses, and on D strike the Arch F, then* 1.12 draw the Tangent CF, then take the nearest distance from the first division of CD, to the Tangent-line CF, which distance shall divide AB* 1.13 into 5 equal parts, as the given divided line CD; as required.

PROP. IX. How to Protract or lay down any of the Regular Figures, called Polygons.

To perform which divide 360 degrees, (the number of degrees in a Circle) by the number of the Poligon his sides: as if it be a Pentagon by 5, if a Hexagon by 6, &c. the Quotient is the Angle of the Center; its Complement to 180D. (or a Semi circle) is the Angle at the Figure, half whereof is the Angle of the Tri∣angle at the Figure: Now I will shew how to delineate any Poligon three ways, viz. 1 by the Angle at the Center, 2. by the Angle at the Figure, 3. by the Angle of the Triangle at the Figure: I have hereunto annexed a Table, which gives at the first sight, (without the trouble of Division) 1. the quantity of the Angle at the Center; 2. the quantity of the Angle at the

Figure; and 3 the Quantity of the Angle at the Triangle of the Figure, from a Triangle to a Decigon.

Names of the Poli∣gons.	Sides	Angles at the Center	Angles at the Figure	Angles at the Trian.
D	M	D	M	D	M
Triangle	3	120	00	60	00	30	00
Square	4	90	00	90	00	45	00
Pentagon	5	72	00	108	00	54	00
Hexagon	6	60	00	120	00	60	00
Heptagon	7	51	43½	128	34½	64	17¼
Octogon	8	45	00	135	00	67	30
Nonigon	9	40	00	140	00	70	00
Decigon	10	36	00	144	00	72	00

CONSTRUCTION I.

First by the Angle at the Center, to delineate a Hexagon, whose Angle at the Center is 60 degrees, first lay down an Angle of 60 deg.* 1.14 (by prop. the 5. aforegoing) making its sides of a convenient length at pleasure, then take such a distance from O the Center of the figure, equally on both sides, as may make the third side equal to the side of the Poligon gi∣ven; which here is 100 parts: * 1.15 Then divide the third side equally into two equal parts, and draw a line through it, from ☉ the Center:

set each half of the side of the Poligon 100, to wit 50, on each from the middle of the third line.† 1.16 thus having placed the side of the Hexagon PP, 100 parts, in order; de∣scribe the whole Hexa∣gon PPPPPP, as was re∣quired.

CONSTRUCTION II.

Now by the Angle of the Figure, to de∣lineate any regular Poligon, Let it be required to protract a Hexagon, whose side as afore is 100 parts; first I draw a line and make it 100 of those parts, then I sind in the precedent Table the Angle of a Hexagon at the figure to be 120 degrees: Then on each side of the drawn line, I lay down an Angle of 120 deg. (according to the 5 precedent propositions) and so work 6 times, (or as many times as your Poligon hath sides) making each side 100 parts,* 1.17 and each Angle 120 degrees; so shall you have enclosed the Poligon PPPPPP, as required.

CONSTRUCTION. III.

To Protract or lay down a Hexagon, or any other regular Poligon, by the Angle of the Triangle, do thus; First draw the side of the* 1.18 Hexagon P P, make it 100 parts. I find in the precedent Table that the Angle of the Triangle is 60 deg; then at each end of the line P P, I

lay down an Angle of 60 * 1.19 deg. (by prop. 5. prece∣dent) and continue the two lines PO, and PO; untill they intersect each* 1.20 other in O: then on O, as a Center (OP: be∣ing Radius) describe a Circle, and within it de∣scribe the Hexagon PPPPPP, as you see in the figure: and so may you delineate any other Poligon: whose Angels from a Triangle, to a Decigon, are all specified in the precedent Table.

PROP. X. To divide a line according to any assigned pro∣portion.

Admit the right line given to be AB, and 'tis required to divide the same into two parts, bearing proportion the one to the other as the lines E, and F doth: To perform which, first draw the line CD, equal to the given line AB: Then draw the line HC, from C, to contain* 1.21 an Angle at pleasure. Then from C to G, place the line F, and from G, to H, place the line E: Then draw the line HD. And lastly, draw GK parallel to HD, (by the 4 prop. precedent) so is the line DC, equal to AB, and divided into two parts, bearing such proporti∣on to each other, as the two given lines E, and F, as was required.

PROP. XI. To two lines given, to find a third proportional to each of them.

Admit the two given lines be A and B, and* 1.22 'tis required to find a third proportional to A, as A, to B: First make an Angle at pleasure; as HIK. Then place the line B, from I, unto P; and the line A, from I, unto L; and draw PL. then also place the line A, from I unto M, and draw QM, parallel unto LP, (by 4 prop.) so shall the line IQ, be a third proportional unto the two given lines A, and B, as was required. For as B, is to A, so is A, unto the proportional found IQ.

PROP. XII. To three lines given to find out a fourth proportio∣nal unto them.

Admit the three given lines to be A, B, and C; and 'tis required to find a third proportional to them, which shall have such proportion unto A, as B, hath unto C. To perform which, first make an Angle at pleasure as DKG, now see∣ing the line C, hath such proportion to B, as the line A, unto the line sought: Therefore place* 1.23 the line C, from K, unto H and B, from K, to F, and draw FH. Again, place the line A, from K, to I, and draw IE, parallel unto FH, (by 4 prop.) until it cutteth DK, in E; so have you the line KE, a fourth proportional, as was required. For as C, is unto B, so is A, unto the found line KE.

PROP. XIII. To find a mean proportional Line between any two right lines given.

Let the two given lines be A, and B, be∣tween which it is required to find a mean pro∣portional line. To perform which, first joyn the two lines A, and B together, so as they make the right line CED: Then describe thereon a* 1.24 Semicircle CFD. Then on the point E, erect the perpendicular EF, (by 1 prop.) to cut the limb of the Semi-circle in F, so shall EF, be a mean proportional line, between the two given lines A, and B, as required.

PROP. XIV. To find two mean proportional Lines between any two right Lines given.

Let the two given lines be A, and B; be∣tween* 1.25 which 'tis required to find two mean proportionals. To per∣form which, first make an* 1.26 Angle containing 90 deg. making the sides CD, and CE of a convenient length: then from C, place the line B, unto F, and the line A, from C, unto G; and draw FG, which divide equally in H, and describe the Se∣mi-circle F K G. Then take the line B in your Compasses, and place∣ing one soot in G, with the other make a mark

in the limb of the Semi-circle in K, then draw ST, in such sort that it may justly touch the Semi-circle in K, and may cut through the two sides of the Angle, equidistant from the Cen∣ter of the Semi-circle H; so shall SF, and TG, be two mean proportionals, betwixt the two gi∣ven lines A, and B, as required.

PROP. XV. To make a Geometrical square equal to divers Geo∣metrical squares.

Let there be given the 5 sides of five Geome∣trical Squares, viz. A, B, C, D, E; and 'tis requi∣red to make one Geometrical Square, equal to the said five Sqares: To perform which first make a Right Angle as ABC, making its contain∣ed* 1.27 sides of a convenient length. Then from B, place A, to D, and from B, place B, to E, and draw Ed. Then place Ed, from B, to F, and C, from B, to G; and draw GF. Then place GF, from B, to H, and D, from B, to I; and draw ••H. Lastly from B, unto K, place IH, and from B, unto L, place the line E; and draw LK. So shall LK, be the side of a Square, equal to the five Squares propounded.

PROP. XVI. To make a Circle equal to divers Circles propounded.

Let the two Circles propounded be A,* 1.28 and B, and 'tis required to make a third Circle, ••e••ual to the said Circles propounded. To per∣form

which, first take the Diameter, of the les∣ser Circle A, and place it as a Tangent, on the Diameter of the greater Circle B, at right An∣gles;* 1.29 as ECD. Then draw the Diagonal ED, which divide equally in F, on which as a Cen∣ter describe the Circle K, making E D, the Diameter of which Circle K shall be equal unto the two given Circles A, and B, as required* 1.30

SECT. II. Of Planometry, or the way to measure any plain Superfice.

PLanometry is that part of the Mathematicks, derived from that Noble Science Geometry, by which the Superficies or Planes of things are measured, and by which their Superficial Content is found, which is done most com∣monly by the Squares of such Measures, Viz. a Square Inch, Square Foot, Square Yard, Square Pace, Square Perch, &c. That is whose side is an Inch, Foot, Yard, Pace, or Pearch Square. So that the Content of any Figure is said to be found, when you know how many such Inches, Feet, Yards, Paces, &c. are contained therein: Thus the End and Scope of Geometry is to measure well.

PROP. I. To find the superficial Content of a Geometrical square.

Let the side of the Square AA be 4 Perch,* 1.31 what is the Area, or superficial content thereof? To find which multiply its side 4, by its self, it produceth 16, which is the content of that Square AAAA, propounded.

PROP. II. To find the superficial content of a Parallelogram, or long Square.

Multiply the length in parts, by the breadth in parts; the product is the content thereof. So in the Parallelogram, or long Square ABCD, the length of the side AB, or CD is 20 Paces,* 1.32 and the breadth AC, or BD is 10 paces, and his superficial content is required. I say there∣fore if according unto the Rule, you multi∣ply the length 20, by the breadth 10, it produ∣ceth 200 Paces; which is the content of the Pa∣rallelogram or long Square ABCD.

PROP. III. To find the superficial Content of any Right-lined Triangle.

Although right-lined Triangles are of seve∣ral kinds, and forms; as first in respect unto their Angles, they are either Right-angled, or

Oblique-angled, i. e. Acute-angled, or Obtuse∣angled. Secondly in respect of their sides, they are either an Equilateral, Isosceles, or Scaleni∣um Triangle: But now seeing they are all measur'd by one and the same manner, I shall therefore add but one Example for all; which take for a general Rule: which is,

Multiply the length of the Base, by the* 1.33 length of the Perpendicular, half their product is the Area or superficial content thereof. So if the content of the Triangle ABC, be required. To find which first from the Angle B, let fall the Perpendicular DB, on the Base AC, (by prop. 3. §. 1.) let therefore the length of the* 1.34 Perpendicular BD be 24, and the Base AC 44 parts. Now if the Base AC 44, were multiply∣ed by BD 24, the product is 1056, half where∣of is 528, the Content of the Triangle ABC, pro∣pounded.

PROP IV. To find the superficial Content of a Rhombus.

First let fall a Perpendicular from one of the Obtuse-angles, unto its opposite side, (by prop. 3. §. 1.) and then Multiply the length of the side thereof, by the length of the Perpendicu∣lar, their product is the Content thereof.

So in the Rhombus ABCD, the side AC, or BD is 16 Inches, and the Perpendicular KC is* 1.35 14 Inches, which multiplyed into the side 16, produceth 224 Inches; which is the Area, or superficial Content, of the Rhombus ABCD, pro∣pounded.

PROP. V. To find the superficial content of a Rhomboides.

Frst let fall a Perpendicular, as in the former proposition, then the length thereof multiply by the length of the Perpendicular; the pro∣duct is the Area, or superficial content thereof. For in the Rhomboides EDAH, whose length AH, or ED is 32 Feet, and the length of the* 1.36 Perpendicular HK is 16 Feet, which multiply∣ed together produceth 512 Feet, which is the Area or superficial content of the Rhomboides AHED, propounded.

PROP. VI. Te find the superficial Content of any Poligon, or many equal sided Superficies.

First from the Center unto the middle of either of the sides of the Poligon, let fall a Per∣pendicular, (by 3. prop §. 1.) Then multiply the length of half the Perifery, by the Perpen∣dicular, the product shall be the Superficial Con∣tent of the Poligon.

Admit the Poligon to be an Hexagon AAAA* 1.37 AA, whose side AA is 22 Feet, and the Per∣••end••cular BE 19 Feet; now, if 66 half the Pe∣rifery, be multiplyed by 19 it produceth 1254 Feet; which is the Content of the Poligon AA, &c. as required.

PROP. VII. To find the superficial Content of a Circle.

Multiply half the Circumference, by one half of the Diameter, their product is the superficial Content thereof.

Admit the Circumference of a Circle ACBD, be 44 Inches, what is the Area or Content thereof. (by the 9. prop. §. 2.) I find the Dia∣meter* 1.38 to be 14 Inches, therefore I say if 22, half the Circumference, be multiplied by 7, half the Diameter, it shall produce 154 Inches; which is the superficial Content of the Circle ACDB, as required.

PROP. VIII. By the Diameter of a Circle given, to find the Cir∣cumference.

Suppose the Diameter be 14, what is the Circumference? The Analogy or Proportion holds thus, as 7, to 22, so is 14, unto 44, the Circumference required.

PROP. IX. By the Circumference of a Circle given, to find the Diameter.

Suppose the Circumference of a Circle be 44 what is the Diameter? the Analogy or Proportion is, as 22, to 7, so is 44, unto 14, the Diameter re∣quired.

Now the proportion of the Diameter, unto the Circumference is as 7, unto-22; or as 113, to 355; or as 1, unto 3, 1415926, &c. so is the Diameter to the Circumference.

PROP. X. By the Content of a Circle given, to find the Cir∣cumference.

Suppose the Content of a Circle be 154, what is the Circumference, the Analogy or Proportion?

As 7, unto 4 times 22, which is 88, so is 154 the Content of the given Circle; to the square of the Circumference 1936, whose root being Extracted, as is taught (in prop. 8. §. 1. chap. 1.) gives the Circumference 44, as required.

PROP. XI. By the Content of a Circle given, to find the Dia∣meter.

Suppose the Superficial Content of a Circle be 154 parts, what is the Diameter thereof? to find which this is the Analogy or Proportion.

As 22,

To 4 times 7, which is 28,

So is 154, the given Content,

To the Square of the Diameter 196, whose Root being Extracted (by 8 prop chap. 1. §. 1.) ••iveth the Diameter 14, as required.

PROP. XII. By the Diameter of a Circle given to find the side of a square equal thereto.

To find which this is the Analogy or Propor∣tion.

As 1, 000000,

To 0, 886227.

So is the Diameter of the Circle propounded.

To the side of a Square, whose superficial Content, is equal unto the superficial Content, of the Circle propounded.

PROP. XIII. By the Circumference of a Circle given, to find the side of a square equal to it.

This is the Analogy or Proportion.

As 1. 000000,

To 0. 282093.

So is the Circumference of the Circle pro∣pounded, to the side of a Square equal to the Circle.

PROP. XIV. By the Content of a Circle given to find the side a square equal to it.

To do which, Extract the Square-Root o•• the Content propounded, (by prop. 8 chap. •••• §. 1.) so is the Root, the side of a Geometrica•• Square, equal thereunto.

PROP. XV. By the Diameter of a Circle given, to find the side of an Inscribed square.

This is the Analogy or Proportion.

As 1. 000000,

To 0. 707107,

So is the Diameter of the Circle propounded, To the side of the inscribed Square.

PROP. XVI. By the Circumference of a Circle given, to find the side of an Inscribed Square.

This is the Analogy, or Proportion.

As 1. 000000,

To 0. 225079.

So is the Circumference of the Circle pro∣pounded, To the side of the inscribed Square.

PROP. XVII. ••o find the Superficial Content of an Oval, or El∣leipsis.

Let the Oval given be ABCD, and 'tis re∣••uired to find the Area or Superficial Content 〈…〉〈…〉ereof? To do which multiply the length A 〈…〉〈…〉 40 Inches, by the breadth CD 30 Inches, the* 1.39 •••• is 1200. Which divide by 1. 27324; 〈…〉〈…〉e Quotient is 942 48/100 parts. Which is the 〈…〉〈…〉ea or Superficial Content of the Oval ABCD 〈…〉〈…〉opounded

PROP. XVIII. To find the Superficial Content of any Section, or Portion of a Circle.

Multiply half the Circute of the Section, by the Semidiameter of the whole Circle, and the product thence arising is the Area or superfi∣cial Content thereof.

Suppose there be a Circle whose Diameter is* 1.40 14 parts, and the Circute of the Quadrent ABC is 11 parts, and the Content of the said Qua∣drent is desired? To find which multiply 5 ½ or, 5., 5 half the Circute of the Quadrent, by 7 the Semidiameter, the product is 38 5/10, which is the Content of the Quadrent ABC pro∣pounded.

SECT. III. Of STEREOMETRY, or the way how to measure any Regular Solid.

STereometry is that part of the Mathema∣ticks, springing from Geometry, by which the Content of all Solid Bodies are discovered by two Multiplications, or three Dimention and is valued by the Cube of some famous Mea sure; as an Inch-Cube, a Foot-Cube, a Yard Cube, or a Perch-Cube, &c.

PROP. I. To find the solid Content of a Cube.

Multiply the side into its self, and that pro∣duct by its side again; their product is the so∣lid Content thereof.

Suppose there be a Cube A, whose side is* 1.41 2 Feet; and his solid Content is required? I say if his side 2, be multiplyed by its self, it pro∣duceth 4, which again multiplyed by 2, it pro∣duceth 8 Feet, which is the solid Content of the Cube propounded.

PROP. II. To find the solid Content of a Parallelepipedon.

First get the Superficial Content of the End, (by prop. 1, or 2, §. 2.) which multiply into the length, the product is the solid Content.

Suppose there be a Parallelepipedon B, whose* 1.42 sides of the Base is 40, and 30 Inches, and length 120 Inches, and his Solid Content is demanded? I say if you multiply 30, by 40, the product is 1, 200, which is the superficial Content at the Base. Which multiplyed by the length 120 In∣ches produceth 144000 Inches, which is the so∣lid Content of the Parallelepipedon B, propounded.

PROP. III. To find the solid Content of a Cylinder.

First get the superficial Content of the Circle at the Base, (by prop. 7. §. 2.) and by it multi∣ply its length, their product is the solid Content thereof.

Suppose there be a Cylinder as D, whose* 1.43 Diameter of the Circle at the Base is 7 parts, and the length of the Cylinder is 14 parts, and 'tis required to find the solid Content thereof? First I find the superficial Content of the Base to be 38. 5, which multiplied into 14 the length, giveth 539 parts, which is the solid Content of the Cylinder propounded.

PROP. IV. To find the solid Content of a Pyramid.

First get the superficial Content of the Base of the Pyramid, (by some of the aforegoing pro∣positions in Planometria) and then multiply that into ⅓ of his Altitude, the product is the solid Content thereof.

Suppose there be a Pyramid H, whose side of the Base is 4 ½ parts, or 4 5/10, and his Altitude 12 parts, and his solid Content is required? First I find, (by prop. 1. §. 2.) the superficial Content* 1.44 of the Base to be 20 25/100 or 20 ¼, which multiply∣ed by 4, (which is ⅓ of the Altitude 12) produ∣ceth 81 parts, for the solid Content of the Py∣ramid propounded.

PROP. V. To find the solid content of a Cone.

First find the superficial Content of the Circle at the Base, (by prop. 7. §. 2.) then multiply it by ⅓ of its Altitude or Heighth, the product is the solid Content thereof.

Suppose there be a Cone as B, whose Dia∣meter* 1.45 of the Base is 7, and his Altitude or Heighth is 15 parts, and his solid Content is re∣quired? First I find the superficial Content of the Base to be 38½ or 38. 5; which multiply∣ed into 5, ⅓ of its Altitude or Heighth) produ∣ceth 192. 5, or ½, which is the solid Content of the Cone propounded.

PROP. VI. By the Diameter of a Globe to find his solid Content.

This is the Analogy or Proportion.

As 6 times 7, which is 42.

Is to 22,

So is the Cube of the Diameter of the Sphere, or Globe propounded.

To the solid Content thereof.

Suppose there be a Sphere or Globe, whose Diameter is 12 Inches; what is the solid Con∣tent thereof? say, (see the Globe R.)

As 42,

Is to 22,* 1.46

So is 1728, the Cube of the Diameter,

To the solid Content 905 6/42 or 1/7 of the Globe, or Sphere propounded: This and all other such* 1.47 like Propositions, are performed by the help of the first Proposition, of the first Chapter of this Book.

PROP. VII. By the Circumference of a Sphere, or Globe, to find his solid Content.

This is the Analogy or Proportion.

As 1. 000000,

To 0. 016887,

So is the Cube of the Circumference of the Globe or Sphere propounded

To the solid Content thereof.

PROP. VIII. By the Axis of a Globe, to make a Cube equal there∣unto.

This is the Analogy or Proportion.

As 1. 00000,

To 0. 80604,

So is the Axis of the Sphere propounded,

To the ••u••••-Root, which shall be equal to it.

PROP. IX. By the Circumference of a Globe, to make a Cube equal thereunto.

This is the Analogy or Proportion.

As 1. 000000,

To 0 256556.

So is the Circumference of the Globe propoun∣ded,

To the Cube-Root, which shall be equal to the Sphere, or Globe, propounded.

PROP. X. By the solid Content of a Sphere or Globe, to make a Cube equal thereunto.

Extract the Cube-root of the solid Content of the Sphere or Globe, (by prop. 9. § 1. chap. 1.) so shall the Root, so found, be the side of a Cube, equal unto the Globe or Sphere propounded.

PROP. XI. A Segment of a Sphere being given to find the solid Content thereof.

To find which first say, As the Altitude of the other Segment, is to the Altitude of the Seg∣ment

given: so is that Altitude of the other Segment increased by half the Axis, unto a fourth: Then say, As 1, to 1, 0472, so is the product of the Quadrant of half the Chord of the Circumfe∣rence of that Segment, multiplyed by that fourth, To the solid Content of the Segment propounded.

Notes

• * 1.1

  Defin.
• * 1.2

  Quod quaeritur cog∣noscendi illius gratia, quod semper est, non & ejus quod oritur, quando∣que & interit. Geometria, ejus quod est semper, Cogni∣tio est. Ac tollet igitur (ô Generose vir) ad veritatem, animum: atque ita, ad Philosophandum prepar••vit cogi∣tationem, ut ad supera convertamus: quae nunc, contra quàm decet, ad inferiora dejicimus, &c. Plato lib. 7. de Rap.
• * 1.3

  Fig. 1,
• * 1.4

  Fig. 2,
• * 1.5

  Fig. 2.
• * 1.6

  Fig. 3.
• * 1.7

  Fig. 4.
• * 1.8

  Fig. 5.
• * 1.9

  Fig. ••.
• * 1.10

  Fig. 6.
• * 1.11

  Fig. 7.
• * 1.12

  Fig. 8.
• * 1.13

  Those two propositions well understood, doth de∣monstrate many other pro∣positions, and thereon is grounded the vse of the Sector.
• * 1.14

  〈…〉〈…〉 9.
• * 1.15

  But this in a Hexa∣gon need not be done, be∣cause the 3 sides of the Triangle are equal, but in all other Poligons it must be done.
• † 1.16

  But if the third line do exceed or be short of the side of the Poligon pro∣pounded, then by parallels on each side, cut the sides of the Triangle, till you have found by those Inter∣sections where to set the line proposed, in any Po∣ligon, &c.
• * 1.17

  Fig. 10.
• * 1.18

  Fig. 11
• * 1.19

  ☞ Observe these Rules well, for you will find them of infinite use in Fortification, &c.
• * 1.20

  Fig. 11.
• * 1.21

  Fig. 12.
• * 1.22

  Fig. 13.
• * 1.23

  Fig. 14.
• * 1.24

  Fig. 15.
• * 1.25

  Fig. 16.
• * 1.26

  And this of all other the Inventions of Plato, Apollonius, Sporus, Archi∣tas, Diocles, Nicomedus; & many other famous Geo∣metricians and Philoso∣phers, I like best for the ready performance of this Conclusion, whose several Methods I could here de∣scribe, but for brevity sake do omit them.
• * 1.27

  Fig. 17.
• * 1.28

  Fig. 18.
• * 1.29

  Fig. 18.
• * 1.30

  So after the same manner, may divers Cir∣cles be added into one by the help of the former proposition well understood.
• * 1.31

  Fig. 19.
• * 1.32

  Fig. 20.
• * 1.33

  The Rule.
• * 1.34

  Fig. 21.
• * 1.35

  Fig. 22.
• * 1.36

  Fig. 23.
• * 1.37

  Fig. 24.
• * 1.38

  Fig. 25.
• * 1.39

  Fig. 26.
• * 1.40

  Fig. 27.
• * 1.41

  Fig. 28.
• * 1.42

  Fig. 29.
• * 1.43

  Fig. 30.
• * 1.44

  Fig. 31.
• * 1.45

  Fig. 32.
• * 1.46

  Fig. 33.
• * 1.47

  ☞ Note that every Sphere is equal unto two Cones, whose Height and Diameter of the Base is the same with the Axis of the Sphere. And a Sphere is two thirds of a Cylinder, whose Height and Diameter of the Base is the same with the Axis of the Sphere; according unto the 9th. Manifestation of the first Book of Archimedes of the Sphere and Cylinder